Question

A random variable X follows a Binomial distribution; i.e. X ~ B (12, p) with a variance

of 1.92. Compute the possible values of p. What is the larger value of p?

Answer 1

The variance of the binomial distribution is given by Var(X) = np(1-p), where n is the number of trials and p is the probability of success in each trial

Here, we are given that X ~ B(12, p) and Var(X) = 1.92. Substituting the values, we get:

Var(X) = np(1-p) = 12p(1-p) = 1.92

Simplifying this equation, we get a quadratic in p:

12p - 12p^2 = 1.92

12p^2 - 12p + 1.92 = 0

Solving for p using the quadratic formula, we get:

p = [12 ± sqrt(12^2 - 4*12*1.92)]/(2*12)

p = [1 ± sqrt(1 - 0.64)]/2

p = [1 ± 0.6]/2

The possible values of p are 0.7 and 0.3.

Since we have two possible values of p, we need to determine which one is larger. We can do this by calculating the mean of the binomial distribution, which is given by np

Using p = 0.7, we get:

mean = np = 12*0.7 = 8.4

Using p = 0.3, we get:

mean = np = 12*0.3 = 3.6

The larger value of p is 0.7